Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int n, q;
long long lft[222222], rgt[222222];
long long pref[222222];
int main() {
  long long sy, a, b;
  scanf("%lld%lld%lld", &sy, &a, &b);
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); (i)++) scanf("%lld%lld", lft + i, rgt + i);
  for (int(i) = 0; (i) < (n); (i)++) pref[i + 1] = pref[i] + rgt[i] - lft[i];
  scanf("%d", &q);
  for (int(i) = 0; (i) < (q); (i)++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    long double L = x == a ? x : (x - y * (x - a) / (long double)(y - sy));
    long double R = x == b ? x : (x - y * (x - b) / (long double)(y - sy));
    int lbin = 0, rbin = n;
    while (lbin + 1 < rbin) {
      int mid = (lbin + rbin) / 2;
      if (L > rgt[mid] - 0.0000000001) {
        lbin = mid + 1;
      } else {
        rbin = mid;
      }
    }
    int anshf = lbin;
    if (anshf == n) {
      printf("%.10lf\n", 0);
      continue;
    }
    lbin = 0, rbin = n;
    while (lbin + 1 < rbin) {
      int mid = (lbin + rbin) / 2;
      if (R < lft[mid] + 0.0000000001) {
        rbin = mid;
      } else {
        lbin = mid;
      }
    }
    long double ratio = 0.0;
    if (anshf == lbin) {
      ratio = max(0.0L, min(R, (long double)rgt[anshf]) -
                            max(L, (long double)lft[anshf]));
      ratio /= R - L;
    } else if (anshf + 10 >= lbin) {
      for (int xx = anshf; xx <= lbin; xx++) {
        ratio += max(
            0.0L, min((long double)rgt[xx], R) - max(L, (long double)lft[xx]));
      }
      ratio /= R - L;
    } else {
      for (int xx = anshf; xx <= anshf + 2; xx++) {
        ratio += max(
            0.0L, min((long double)rgt[xx], R) - max(L, (long double)lft[xx]));
      }
      for (int xx = lbin - 2; xx <= lbin; xx++) {
        ratio += max(
            0.0L, min((long double)rgt[xx], R) - max(L, (long double)lft[xx]));
      }
      ratio += pref[lbin - 2] - pref[anshf + 3];
      ratio /= R - L;
    }
    ratio = max(0.0L, ratio);
    printf("%.10lf\n", (double)(ratio * (b - a)));
  }
  return 0;
}